1995 AIME Problem 9

Attempt Problem 9 of the 1995 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AIME solutions, or check the answer key.

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9.

Triangle

ABC

is isosceles, with

AB=AC

and altitude

AM=11.

Suppose that there is a point

D

on

\overline{AM}

with

AD=10

and

\angle BDC=3\angle BAC.

Then the perimeter of

\triangle ABC

may be written in the form

a+\sqrt b,

where

a

and

b

are integers. Find

a+b.

Answer

Answer: 616

Concepts:trigonometric identity isosceles triangle trigonometry

Difficulty rating: 2170

Small hint:

Let

BM=x

and let half the apex angle be

\alpha

Big hint:

Use

\tan\alpha=\frac{x}{11},

DM=1,

and the triple-angle formula for tangent

Solution:

Let

BM=x

and

\alpha=\angle BAM,

so

\tan\alpha=\frac{x}{11}.

Because

DM=AM-AD=1,

symmetry gives

\angle BDC=2\arctan x,

while

\angle BAC=2\alpha.

Hence

\arctan x=3\alpha.

Put

t=\tan\alpha=\frac{x}{11}.

Then

\frac{3t-t^3}{1-3t^2}=11t,

so

t^2=\frac{1}{4}

and

x=\frac{11}{2}.

Thus

BC=11

and

AB=\frac{11\sqrt5}{2},

making the perimeter

11+11\sqrt5=11+\sqrt{605}.

Therefore

a+b=11+605=616.